Question

Find the \(n^{\text {th }}\) term of the indicated Taylor polynomial. Find a formula for the \(n^{\text {th }}\) term of the Maclaurin polynomial for \(f(x)=\sin x\).

Short answer

The nth term is 0 if \(n \equiv 0, 2 \,(\text{mod 4})\); \(\frac{x^n}{n!}\) if \(n \equiv 1 (\text{mod 4})\); \(-\frac{x^n}{n!}\) if \(n \equiv 3 (\text{mod 4})\).

Step-by-step solution

Identify the Function and Maclaurin Polynomial Formula

The function given is \(f(x) = \sin x\). A Maclaurin polynomial is a Taylor series centered at 0. The formula for the \(n^{\text{th}}\) term of a Taylor polynomial at \(x = 0\) (Maclaurin series) is: \[ T_n(x) = \frac{f^{(n)}(0)}{n!}x^n \] where \(f^{(n)}(0)\) is the \(n^{\text{th}}\) derivative of \(f(x)\) evaluated at 0.

Compute the Derivatives of the Function

Calculate the first few derivatives of \(f(x)=\sin x\) to find a pattern: - \(f(x) = \sin x\)- \(f'(x) = \cos x\)- \(f''(x) = -\sin x\)- \(f'''(x) = -\cos x\)- \(f^{(4)}(x) = \sin x\)These derivatives repeat every four terms: \(\sin x, \cos x, -\sin x, -\cos x, \sin x, \ldots\).

Evaluate the Derivatives at Zero

Now, we'll evaluate these derivatives at \(x = 0\):- \(f(0) = \sin(0) = 0\)- \(f'(0) = \cos(0) = 1\)- \(f''(0) = -\sin(0) = 0\)- \(f'''(0) = -\cos(0) = -1\)- \(f^{(4)}(0) = \sin(0) = 0\)The evaluations repeat: \(0, 1, 0, -1, 0, \ldots\).

Write the General Formula for the n-th Term

The \(n^{\text{th}}\) term of the Maclaurin series takes the form depending on the pattern of the derivatives evaluated at zero:If \(n = 4k\) or \(n = 4k+2\), the \(n^{\text{th}}\) term is 0.If \(n = 4k+1\), the \(n^{\text{th}}\) term is \(\frac{x^n}{n!}\).If \(n = 4k+3\), the \(n^{\text{th}}\) term is \(-\frac{x^n}{n!}\).

Conclude with the Maclaurin Polynomial Formula

Thus, the formula for the \(n^{\text{th}}\) term of the Maclaurin series for \(\sin x\) is:- 0 if \(n\equiv 0 \) or \(n\equiv 2 \mod 4 \)- \(\frac{x^n}{n!}\) if \(n\equiv 1 \mod 4 \)- \(-\frac{x^n}{n!}\) if \(n\equiv 3 \mod 4 \)

Key concepts

Taylor series

A Taylor series is a way to represent a function as an infinite sum of terms. Each term of the series is derived from the derivatives of the function at a single point.

This point is usually chosen based on the problem to simplify calculations. The general formula for the Taylor series of a function, centered at point \(a\), is given by:\[T(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots\]Here, \(f^{(n)}(a)\) denotes the \(n^{\text{th}}\) derivative of the function evaluated at \(a\). Each derivative is divided by \(n!\) (n-factorial) to ensure the series converges to the function itself.

The Taylor series is particularly useful because it provides polynomial approximations to functions that might otherwise be too complex to work with directly. For example, the Taylor series allows us to approximate calculus functions like \(e^x\), \(\sin x\), and \(\cos x\) using polynomial terms.

• Useful for making functions easier to manipulate
• Centered around a point \(a\) for simple evaluations

derivatives

Derivatives are fundamental to understanding the Taylor series. They provide the backbone needed to construct each term of a Taylor polynomial. Derivatives measure how a function changes as its input changes. For example, the derivative of \(f(x) = \sin x\) gives us \(f'(x) = \cos x\), which explains how \(\sin x\) changes to \(\cos x\) when \(x\) increases.

Derivatives are not only about finding slopes; they're also used in approximations and optimizations, and are crucial in physics to describe motion.

For building a Taylor series, we often compute several derivatives, evaluating them at a specific point such as zero for a Maclaurin series:

• First derivative: gives the rate of change of the function
• Second derivative: the rate of change of the rate of change
• The pattern helps in crafting the series and looking for patterns

These patterns help predict the \(n^{\text{th}}\) term, which is crucial in creating a polynomial that approximates the function.

polynomial

Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. They are written in a form such as \(ax^n + bx^{n-1} + ... + cx + d\), where \(a, b, c,\) etc., are the coefficients and \(n\) is a non-negative integer.

In the context of the Taylor series, particularly the Maclaurin series, we use polynomials to approximate functions. These polynomials allow complex functions to be expressed as a sum of terms, each involving powers of \(x\). Each term is derived from the derivatives of the function at a certain point.

The polynomial form is valuable because:

• They are easier to differentiate and integrate
• They can efficiently approximate functions near the chosen point

This approximation becomes highly effective when the derivatives follow a particular pattern, as seen in functions like \(\sin x\). At the core, polynomials simplify complex functions into a series of easy to compute terms.

sine function

The sine function, \(\sin x\), is one of the basic trigonometric functions. Trigonometry deals with the relationships between the angles and lengths of triangles. The sine function itself provides the ratio of the opposite side to the hypotenuse in a right triangle.

When it comes to calculus, \(\sin x\) is particularly interesting due to its derivatives. Its behavior repeats every few terms, which is unique and simplifies calculations when constructing Taylor and Maclaurin series.

• The derivative of \(\sin x\) is \(\cos x\)
• The second derivative is \(-\sin x\)
• The third derivative is \(-\cos x\)

This repeating cycle of derivatives happens every four terms, making it easier to develop series representations like the Maclaurin series.

In the Taylor (Maclaurin) series context, evaluating the sine function at zero yields a progression of terms that reflect specific patterns. This enables the creation of a succinct polynomial approximation: